Evaluate :(i)sqrt(3-2sqrt(2))""(ii)sqrt(9-6sqrt(2))

1.	Evaluate :(i)sqrt(3-2sqrt(2))""(ii)sqrt(9-6sqrt(2))
Answer» Solution :(i) Let `SQRT(3-2sqrt(2))=sqrt(x)-sqrt(y)` Squaring both sides, we get. . .(1) `3-2sqrt(2)=(sqrt(x)-sqrt(y))^(2)` `rArr3-2sqrt(2)=x+y-2sqrt(xy)` Comparing rational and IRRATIONAL PARTS on both sides, we get `x+y=3,xy=2` `"Now "(x-y)^(2)=(x+y)^(2)-4xy` =9-8=1(identity) `:.x-y=+-1` `"If " x-y=1`. . .(2) `andul (x+y=3)` Adding, 2x=4 `rArrx=2` `2" "y=1" "rArr" "y=1` `:.sqrt(3-2sqrt(2))=sqrt(2)-sqrt(1)=sqrt(2)-1` `"If "x-y=-1` `andul(x+y=3)`. . . .(3) Adding 2x=2 `:.""x=1` Put x = 1 in equation (3), `1+y=3" "rArr" "y=2` `:." "sqrt(3-2sqrt(2))=sqrt(1)-sqrt(2)=1-sqrt(2)` But it is a negative quantity and SQUARE root `''sqrt''` cannot give a negative value. So, `sqrt(3-2sqrt(2))!=1-sqrt(2)` `"Hence,"sqrt(3-2sqrt(2))=sqrt(2)-1` (II) Let `sqrt(9+6sqrt(2))=sqrt(x)+sqrt(y)` Squaring both sides, we get `:." "9+6sqrt(2)=x+y+2sqrt(xy)` `rArr9+2(3sqrt(2))=x+y+2sqrt(xy)` Comparing, rational and irrational parts on both sides. `:." "x+y=9,sqrt(xy)=3sqrt(2)" "rArr" "xy=18` `"Now, "(x-y)^(2)=(x+y)^(2)-4xy`(identity) =81-72=9 `:." "x-y=+-3` `"If "x-y=3`. . .(2) `andul(x+y=9)` Adding, `2x=12" "rArr" "y=3` `:.sqrt(9+6sqrt(2))=sqrt(6)+sqrt(3)` `:.sqrt(9+6sqrt(2))=sqrt(6)+sqrt(3)` `"If "x-y=-3`. . .(3) `andul(x+y=9)` Adding, `2x=6" "rArr" "x=3` Put x = 3 in (3), we get `3-y=-3" "rArr" "y=6` `:.sqrt(9+6sqrt(2))=sqrt(3)+sqrt(6)`

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